probability

Question

Given 20 people, what is the probability that,
among the 12 months in the year, there are 4
months containing exactly 2 birthdays and 4 containing
exactly 3 birthdays?

Answer 1

Suppose all 20 births are equally likely to be in any of the 12 months of the year and  all are independent.

Probability that  JAN,FEB,MAR,APR have 2 birthdays and MAY,JUN,JUL,AUG have 3 birthdays  and other months have 0 months =

=1/$12^{20}$   *  20!/($(2!)^{4}*(3!)^{4}*(0!)^{4}$   $\approx$ 3 * 10$^{-8}$

Now other ways in which we can choose months that contain 2 birthday or 3 birthdays or 0 birthdays=12!/(4!)$^{3}$

So final probability=3 * 10$^{-8}$ * 12!/(4!)$^{3}$=0.001

Comments

" 4 months containing exactly 2 birthdays and 4 containing exactly 3 birthdays "

means last 4 months containing 15 birthdays

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D order has not been mentioned though whether the 1st four months, 2nd group of 4 months or d last 3rd group of 4 months. ?????

How abt d year being a leap??

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@ Srestha, den how shud it be worked out?? Say. Coz d following link gives d ans as above. :)

 https://www.quora.com/Given-20-people-what-is-the-probability-that-among-the-12-months-of-the-year-there-are-4-months-with-exactly-2-birthdays-and-4-months-with-exactly-3-birthdays

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if we know 5 people birthday, then remaining must also have a birthday.

So, it should be 15 birthday in remaining 4 months.

I donot think , in this case I am wrong

See this link

https://www.khanacademy.org/math/precalculus/prob-comb/prob-combinatorics-precalc/v/birthday-probability-problem